A square planar complex is formed by hybridization :

\(s,\, \ p_x,\, \ p_y,\, \ d_{x^2 - y^2}\)

The correct answer is option 4. \(s,\, \ p_x,\, \ p_y,\, \ d_{x^2 - y^2}\).

Let us dive into the details of square planar hybridization and how it relates to the formation of a square planar complex.

Square Planar Geometry and Hybridization

Square Planar Geometry: In a square planar geometry, the central atom or ion is surrounded by four ligands arranged at the corners of a square, with the metal ion at the center. This arrangement results in a 90° bond angle between adjacent ligands

Hybridization: For a complex to adopt a square planar geometry, the central metal atom or ion must undergo hybridization that results in four equivalent hybrid orbitals lying in the same plane.

Hybridization Process:

Orbitals Involved: The central metal atom or ion typically uses its \(s\) orbital, two \(p\) orbitals (usually \(p_x\) and \(p_y\)), and one \(d\) orbital (specifically \(d_{x^2 - y^2}\)).

Formation of Hybrid Orbitals:

\(s\) Orbital: Contributes one orbital.

Two \(p\) Orbitals: Usually, \(p_x\) and \(p_y\) are involved in hybridization. These orbitals lie in the plane and are orthogonal to each other.

One \(d\) Orbital: The \(d_{x^2 - y^2}\) orbital is oriented in the plane of the ligands and is used for hybridization. It is important because it directly participates in bonding with the ligands.

Hybridization Scheme:

The combination of the \(s\), \(p_x\), \(p_y\), and \(d_{x^2 - y^2}\) orbitals results in four equivalent **\(sp^2d\)** hybrid orbitals. These four orbitals arrange themselves in a square planar configuration around the central metal atom or ion, ensuring that the bond angles are 90°.

Why Other Options Are Incorrect:

Option 1: \(s, p_x, p_y, p_z\)

This would suggest an **\(sp^3\)** hybridization, which leads to a tetrahedral geometry, not square planar.

Option 2: \(s, p_x, p_y, d^{z^2}\)

The \(d_{z^2}\) orbital is oriented along the z-axis, not in the plane required for square planar geometry. This combination would not form a square planar shape.

Option 3: \(s, p_x, p_z, d_{xy}\)

The \(d_{xy}\) orbital is oriented in the xy-plane but is not suitable for square planar geometry, which requires the \(d_{x^2 - y^2}\) orbital.

In summary, the square planar complex formation involves **\(sp^2d\)** hybridization with the involvement of the \(s\), \(p_x\), \(p_y\), and \(d_{x^2 - y^2}\) orbitals. This specific set of orbitals creates four equivalent hybrid orbitals arranged in a square planar geometry, ensuring optimal bonding with ligands and the correct bond angles.